Question

Consider a particle with mass \(m\) moving in a potential \(U=k x^{2} / 2,\) as in a mass-spring system. The total energy of the particle is \(E=p^{2} / 2 m+k x^{2} / 2\). Assume that \(p\) and \(x\) are approximately related by the Heisenberg uncertainty principle, so \(p x \approx h .\) (a) Calculate the minimum possible value of the energy \(E,\) and the value of \(x\) that gives this minimum \(E\). This lowest possible energy, which is not zero, is called the zero-point energy. (b) For the \(x\) calculated in part (a), what is the ratio of the kinetic to the potential energy of the particle?

Answer 1

The particle of mass \(m\)

The particle is moving with a potential energy \((\mathrm{U})=\frac{1}{2} k x^{2}\)

The total energy of the particle \((E)=\frac{p^{2}}{2 m}+\frac{1}{2} k x^{2}\)

The Heisenberg uncertainty principle is

\(p x=h\)

$$ p=\frac{h}{x} $$

Therefore, \(E=\frac{\mathrm{p}^{2}}{2 m}+\frac{1}{2} k x^{2}\)

$$ =\frac{h^{2}}{2 m x^{2}}+\frac{1}{2} k x^{2} $$

The minimum energy exists at \(x\) when \(\frac{d E}{d x}=0\)

\(\frac{d}{d x}\left[\frac{h^{2}}{2 m x^{2}}+\frac{1}{2} k x^{2}\right]=0\)

\(\frac{-2 h^{2}}{2 m x^{3}}+\frac{1}{2} k \times 2 x=0\)

\(k x=\frac{h^{2}}{m x^{3}}\)

\(x^{4}=\frac{h^{2}}{m k}\)

\(x=\frac{\sqrt{h}}{(m k)^{\frac{1}{4}}}\)

Therefore, minimum energy exists at \(x=\frac{\sqrt{h}}{(m k)^{\frac{1}{4}}}\)

The minimum energy of zero-point energy is

\(\frac{h \sqrt{m k}}{2 m}+\frac{1}{2} \frac{k h}{\sqrt{m k}}\)

(b) The kinetic energy \((K)=\frac{p^{2}}{2 m}\)

\(=\frac{h^{2}}{2 m x^{2}}\)

\(=\frac{h^{2}}{2 m h} \sqrt{m k}\)

\(=\frac{h}{2} \sqrt{\frac{k}{m}}\)

The potential energy is \((U)=\frac{1}{2} k x^{2}\)

\(=\frac{1}{2} k \frac{h}{\sqrt{m k}}\)

\(=\frac{h}{2} \sqrt{\frac{k}{m}}\)

Therefore, the ratio of kinetic energy to potential energy is

\(\frac{K}{U}=1\)

Both energies are the same.